Solve for $q$, $ -\dfrac{1}{20q^2} = -\dfrac{4q + 6}{5q^2} + \dfrac{7}{15q^2} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $20q^2$ $5q^2$ and $15q^2$ The common denominator is $60q^2$ To get $60q^2$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{1}{20q^2} \times \dfrac{3}{3} = -\dfrac{3}{60q^2} $ To get $60q^2$ in the denominator of the second term, multiply it by $\frac{12}{12}$ $ -\dfrac{4q + 6}{5q^2} \times \dfrac{12}{12} = -\dfrac{48q + 72}{60q^2} $ To get $60q^2$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{7}{15q^2} \times \dfrac{4}{4} = \dfrac{28}{60q^2} $ This give us: $ -\dfrac{3}{60q^2} = -\dfrac{48q + 72}{60q^2} + \dfrac{28}{60q^2} $ If we multiply both sides of the equation by $60q^2$ , we get: $ -3 = -48q - 72 + 28$ $ -3 = -48q - 44$ $ 41 = -48q $ $ q = -\dfrac{41}{48}$